# de Rham cohomology of smooth affine varieties

Let $U$ be a smooth variety over $\mathbb{C}$. We know that there exists a smooth compactification $X$ such that $X-U$ is a normal crossings divisor $D$ and that the de Rham cohomology of $U$ can be computed using the complex of sheaves with logarithmic differentials on $X$: $$H_{dR}^\ast(U)=\mathbb{H}^\ast(X, \Omega^\bullet_X(\log D))$$

In general, one needs hypercohomology here but I am wondering if, when $U$ is affine, one simply has $$H^\ast_{dR}(U)=H^\ast(\Gamma(\Omega^\bullet_X(\log D))$$ I know that this is true if one uses instead the de Rham complex $\Omega^\bullet_U$ to compute cohomology and I know that $j_\ast \Omega_U^\bullet$ and $\Omega_X^\bullet(\log D)$ are quasi-isomorphic (here $j: U \hookrightarrow X$). Is this enough to conclude?

Unfortunately, no. Let $X$ be an elliptic curve, $D=p$ a point, $U=X-D$ then one sees from topology that $\dim H^1_{dR}(U)=2$. However, your proposed answer gives $$H^1(\Gamma(\mathcal{O}_X)\to \Gamma(\Omega_X^1(\log D)) = H^1(\mathbb{C} \stackrel{0}{\to} \mathbb{C})=\mathbb{C}$$ To understand what goes wrong, note from the spectral sequence (is this OK?) we get another term $H^1(\mathcal{O}_X)$ which needs to be there.

Here are a few additional comments, which might help. Given a bounded complex $\mathcal{F}^\bullet$, there is a spectral sequence $$E^{pq}= H^q(T,\mathcal{F}^p)\Rightarrow \mathbb{H}^{p+q}(T,\mathcal{F}^\bullet)$$

1. When $T=U$, and the sheaves are $\Omega_U^*$, everything is concentrated on $q=0$ line by Serre, so we get the elementary description.
2. When $T=X$ and the sheaves are $\Omega_X(\log D)^*$, we don't have such a vanishing. On the other hand, the spectral sequence degenerates at $E_1$ (Deligne), so you still don't need to do "serious" hypercohomology.
• Oh, thanks for the example... I should have tried that one! So there is no way to avoid hypercohomology in the affine case? – learner92 Jan 29 '15 at 14:56
• Not really, but this is not a bad thing. I added some further comments. – Donu Arapura Jan 29 '15 at 15:21

The de Rham cohomology of an affine variety $U$ is simply $H^\bullet(\Gamma(U,\Omega_U^\bullet))$ where $\Omega_U^\bullet$ is the sheaf of algebraic forms. See Grothendieck (1966), « On the de Rham cohomology of algebraic varieties », Inst. Hautes Études Sci. Publ. Math., (29), p. 96.

In general, you cannot get rid of hypercohomology AND use logarithmic forms only.

• It might be worth pointing out that $\Omega^p_U$ is a coherent sheaf on $U$ (hence has no cohomology if $U$ is affine). However, when one computes on a smooth compactification $X$ (with $D=X\setminus U$ a divisor with strict normal crossings), one has to compute hypercohomology of a complex of coherent sheaves on $X$, which is no more affine, and the spectral sequence has no reason to be trivial (and in general, it isn't, see Donu Arapura's example). – ACL Jan 29 '15 at 18:50